等比数列{a
n}中,已知对任意自然数n,a
1+a
2+a
3+…+a
n=2
n-1,则a
12+a
22+a
32+…+a
n2=( )
A. (2
n-1)
2B.
(2n−1)C. 4
n-1
D.
(4n−1)
人气:375 ℃ 时间:2019-09-30 06:24:47
解答
设等比数列的公比为q,则由等比数列的性质可知数列{an2}是以q2为公比的等比数列Sn=a1+a2+…+an=2n-1∵a1=S1=1,an=Sn-Sn-1=2n-1-(2n-1-1)=2n-1适合n=1∴an=2n−1,则由等比数列的性质可知数列{an2}是以q2=4为公比...
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